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3 years ago

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Suppose that 0.88g of water condenses on a 75.0-g block of iron that is initially at 22 degrees Celsius. If the heat released du

ring condensation goes only to warming the block, what is the final temperature (in degrees Celsius) of the iron block? Assume a constant enthalpy of vaporization for water of 44.0 kJ/mol.

1 answer:

vitfil [10]3 years ago

8 0

Answer:

The final temperature (in degrees Celsius) of the iron block is 85.82 °C

Explanation:

<u>Determine the heat emitted by the condensing water.</u>

qw = 0.88 g H2O $\frac{1 mol H2O}{18.0152 g H2O}$ $\frac{- 44 kJ}{1 mol H2O}$

qw = -2.1492 kJ

<u>Determine the heat absorbed by the iron block</u>

qw = - qw = - (-2.1492 kJ) = 2.1492 kJ

<u>Look up the specific heat for iron. Use the specific heat equation to solve for the final temperature of the iron block.</u>

q = mcΔT

ΔT = q / mc

Temp f = (q / mc) + Temp i

Temp f = 2.1492 x 10^3 J / 75 g (0.449 J/g.°C) + 22 °C

Temp f = 85.82 °C

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Read 2 more answers

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

$\Rightarrow T_{2}=260K$

$\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}$

So at -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

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